Q31 of 141 Page 447

In the given figure, O is the centre of a circle in which ∠AOC = 100°. Side AB of quad. OABC has been produced to D. Then, ∠CBD = ?

Given: ∠AOC = 100°


Here,


(Exterior ∠AOC) = 360° – (interior ∠AOC)


(Exterior ∠AOC) = 360° – 100°


(Exterior ∠AOC) = 260°


We know that,


(Exterior ∠AOC) = 2 × ∠ADC


260° = 2 × ∠ABC


∠ABC = = 130°


∴ ∠ABC = 130°


Here,


∠ABD = ∠ABC + ∠CBD


180° = 130° + ∠CBD


∠CBD = 180° – 130° = 50°


∴ ∠CBD = 50°

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